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In an answer to this question, it has been suggested to consider the following: $$(I+A)(\sum_{j=0}^n(-A)^j)$$

Through a series of algebraic operations, it can be shown that $\sum_{j=0}^n(-A)^j$ is in fact the inverse of $I+A$.

How would we have known to multiply by $\sum_{j=0}^n(-A)^j$? If there isn't an identity or formula that would indicate such a multiplication is a reasonable avenue of inquiry, then how would we otherwise derive $\sum_{j=0}^n(-A)^j$?

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    $\begingroup$ Just think that multiplying by some kind of sum of the powers of $\;A\;$ could help since $\;A^n=0\;$ for some natural $\;n\;$ ... For example, if $\;A^2=0\;$ , then $\;(I+A)(I-A)=I-A^2=I\;$ . This and, of course, the fact that $\;I\;$ commutes with any matrix (of the same order) makes things very simple. $\endgroup$ – DonAntonio Jul 20 at 17:30
  • $\begingroup$ To come up with the proof, it's best to start with a different (more structural) proof: Working in the quotient ring $\mathbb{Z}\left[A\right]/\left(I+A\right)$, we have $I \equiv -A \mod I+A$, so that $I$ is nilpotent modulo $I+A$ (since $A$ is nilpotent, and thus so is $-A$); but this means that $I^k \equiv 0 \mod I+A$ for some $k$, and therefore $I \equiv 0 \mod I+A$, and thus $I+A$ is invertible (since $1 = 0$ only holds in trivial rings). Now, unravel the use of quotient rings and congruences in this proof, extracting an explicit (or recursive) formula for the inverse. $\endgroup$ – darij grinberg Jul 20 at 20:44
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If $z\in\mathbb C$ and $\lvert z\rvert<1$, then$$\frac1{1+z}=1-z+z^2-z^3+\cdots$$In other words,$$(1+z)\left(1-z+z^2-z^3+\cdots\right)=1.$$This, together with the fact that $A^n=0$ if $n\gg1$, should make you think that it would be a good idea to try to prove that an inverse of $\operatorname{Id}+A$ is $1-A+A^2-A^3+\cdots$ (which is a finite sum, in this case).

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I guess a natural way of thinking about this stems from the formula for a geometric series for real numbers: if $|x| < 1$, then \begin{align} \sum_{j=0}^\infty x^j = \dfrac{1}{1-x} \end{align} Now, replacing $x$ with $-x$ gives the formula \begin{align} (1+x)^{-1} = \dfrac{1}{1+x} = \sum_{j=0}^{\infty} (-x)^j \end{align} So, it might be natural to try this for matrices as well. And now if $A$ is nilpotent, then on the RHS, we only have a finite number of summands (in fact at most $n$ summands if $A$ is $n \times n$). Hence, one might expect that $\sum_{j=0}^n (-A)^j$ is the inverse matrix of $I+A$. Then, a simple computation verifies that it actually is.

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Recall that $$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x^2y^{2n-2}-..+y^{2n})$$

Now, since $I$ and $A$ commute, the same formula holds for $x=I$ and $y=A$. Therefore, $$I+A^{2n+1}=(I+A)(I-A+A^2-A^{3}+...+A^{2n})$$ Now use the fact that $A^n=0$ implies $$A^{n}=A^{n+1}=...=A^{2n+1}=0$$

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